$f(t)=-(t-2)(t-15)$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Solution: $\begin{aligned} -(t-2)(t-15)&=0 \\\\ t-2=0&\text{ or }t-15=0 \\\\ t={2}&\text{ or }t={15} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({2})+({15})}{2} \\\\ &={\dfrac{17}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $f\left({\dfrac{17}{2}}\right)$ : $\begin{aligned} f\left({\dfrac{17}{2}}\right)&=-\left({\dfrac{17}{2}}-2\right)\left({\dfrac{17}{2}}-15\right) \\\\ &=-\left(\dfrac{13}{2}\right)\left(-\dfrac{13}{2}\right) \\\\ &=\dfrac{169}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=2 \\\\ \text{larger }t&=15 \end{aligned}$ The vertex of the parabola is at $\left(\dfrac{17}{2},\dfrac{169}{4}\right)$